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Paladinen [302]
3 years ago
11

1/3(x - 10) = - 4 x = ?

Mathematics
2 answers:
ki77a [65]3 years ago
7 0

Answer:

x = -2

Step-by-step explanation:

1/3(x - 10) = - 4

1/3x - 10/3 = - 4

1/3x = -4 + 10/3

1/3x = -2/3

x = -2/3 · 3

x = -2

IrinaK [193]3 years ago
5 0

Answer:

x=-2

Step-by-step explanation:

First use distributive property

1/3*x and 1/3*-10

=1/3x-10/3=-4

Then add 10/3 on both sides so x can be by itself

=1/3x=-4+10/3

Then add -4 and 10/3

=1/3x=-2/3

Then multiply each side by the reciprocal of 1/3 which is 3/1

x=-2

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The answer is 2/16 if you do the math with the numerators only.
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3 years ago
Rogers Company reported net income of $35,000 for the year. During the year, accounts receivable increased by $7,000, accounts p
storchak [24]

Answer:

correct answer is  option C.

Step-by-step explanation:

net income of a year =  $35,000

accounts receivable is increased (AR)= $7,000

accounts payable decrease(AP) = $3,000

depreciation expense = $8,000

net cash provided = net income - increase in current asset (AR) - decrease in  current asset (AP) +non cash flow

net cash provided =  $35,000 -$7,000-$3,000+ $8,000

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3 years ago
(c). It is well known that the rate of flow can be found by measuring the volume of blood that flows past a point in a given tim
aleksklad [387]

(i) Given that

V(R) = \displaystyle \int_0^R 2\pi K(R^2r-r^3) \, dr

when R = 0.30 cm and v = (0.30 - 3.33r²) cm/s (which additionally tells us to take K = 1), then

V(0.30) = \displaystyle \int_0^{0.30} 2\pi \left(0.30-3.33r^2\right)r \, dr \approx \boxed{0.0425}

and this is a volume so it must be reported with units of cm³.

In Mathematica, you can first define the velocity function with

v[r_] := 0.30 - 3.33r^2

and additionally define the volume function with

V[R_] := Integrate[2 Pi v[r] r, {r, 0, R}]

Then get the desired volume by running V[0.30].

(ii) In full, the volume function is

\displaystyle \int_0^R 2\pi K(R^2-r^2)r \, dr

Compute the integral:

V(R) = \displaystyle \int_0^R 2\pi K(R^2-r^2)r \, dr

V(R) = \displaystyle 2\pi K \int_0^R (R^2r-r^3) \, dr

V(R) = \displaystyle 2\pi K \left(\frac12 R^2r^2 - \frac14 r^4\right)\bigg_0^R

V(R) = \displaystyle 2\pi K \left(\frac{R^4}2- \frac{R^4}4\right)

V(R) = \displaystyle \boxed{\frac{\pi KR^4}2}

In M, redefine the velocity function as

v[r_] := k*(R^2 - r^2)

(you can't use capital K because it's reserved for a built-in function)

Then run

Integrate[2 Pi v[r] r, {r, 0, R}]

This may take a little longer to compute than expected because M tries to generate a result to cover all cases (it doesn't automatically know that R is a real number, for instance). You can make it run faster by including the Assumptions option, as with

Integrate[2 Pi v[r] r, {r, 0, R}, Assumptions -> R > 0]

which ensures that R is positive, and moreover a real number.

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